使用消息队列的线程同步不按顺序

Question

以下代码来自Raymond's Pycon keynotes on cuncurrency，非常棒的介绍。有点长，先从问题说起：

如果消息一起添加到队列，并且应该在线程被认为完成之前打印它们，他们怎么会印刷混乱？在线程初始化之间添加 100 毫秒的延迟解决了这个问题，并且输出符合预期。

-

import Queue, time

counter_queue = Queue.Queue()
counter = 0

def counter_manager():
    # I have EXCLUSIVE rights to update the counter variable
    global counter

    while True:
        increment = counter_queue.get()
        counter += 1
        print_queue.put(['The count is {}\n'.format(counter),
                         '----------\n'])
        counter_queue.task_done()

t = threading.Thread(target=counter_manager)
t.daemon = True
t.start()
del t


print_queue = Queue.Queue()

def print_manager():
    # I have EXCLUSIVE rights to call the print keyword
    while True:
        job = print_queue.get()
        for line in job:
            print line
        print_queue.task_done()

t = threading.Thread(target=print_manager)
t.daemon = True
t.start()
del t

def worker_threads():
    counter_queue.put(1)

print_queue.put(['Starting up.. with message queues'])
worker_threads = []
for i in range(10):
    t = threading.Thread(target=worker)
    worker_threads.append(t)
    t.start()
    # time.sleep(0.1)
for t in worker_threads:
    t.join()

counter_queue.join()
print_queue.put(['Finishing up'])
print_queue.join()

它使用counter_manager和print_queue作为守护进程从工作线程接收消息，并顺序执行它们。据我了解，这应该保留执行顺序。但是，我得到以下输出：

Starting up.. with message queues
The couns is 1The couns is 2

--------------------

The couns is 3
 ----------The couns is 4
 The couns is 5
The couns is 6
----------
----------
----------
The couns is 7

----------The couns is 8

The couns is 9----------

----------The couns is 10

----------
Finishing up

计数器正确递增，但打印消息仍然混乱。

如果我取消注释睡眠语句，线程初始化会延迟 100 毫秒并且输出是正确的

Starting up.. with message queues
 The couns is 1
----------
The couns is 2
----------
The couns is 3
----------
The couns is 4
----------
The couns is 5
----------
The couns is 6
----------
The couns is 7
----------
The couns is 8
----------
The couns is 9
----------
The couns is 10
----------
Finishing up

使用队列，打印应该是有序的。这不对吗？

工人代码

def worker():
    global counter

    counter += 1
    print 'The couns is {}'.format(counter)  # read the var (race cond.)
    print '----------'

Answer 1

我无法重现这个问题。为了达到运行，我添加了一个 threading 导入并将 worker_threads 函数重命名为 worker。

输出有 "couns" 但您的代码显示 "count"。您能否验证您提供的输出和代码是否一致？

您的输出似乎只在行级别交错。

我想知道以前版本的代码是否在 counter_manager 中有两个不同的 .put（不是原子的）后来合并成一个多行的 .put？

使用消息队列的线程同步不按顺序

thread synchronization using message queues is not in order

python

multithreading

message-queue

python-multithreading

python-2.7